Definite Integral from 0 to 2 Pi of Reciprocal of One minus 2 a Cosine x plus a Squared

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Theorem

$\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2} = \frac {2 \pi} {1 - a^2}$

where $a$ is a real number with $0 < a < 1$.


Proof



Let $C$ be the unit open disk centred at $0$.

The boundary of $C$, $\partial C$, can be parameterized by:

$\map \gamma \theta = e^{i \theta}$

for $0 \le \theta \le 2 \pi$.

We have:

\(\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2}\) \(=\) \(\ds \int_0^{2 \pi} \frac {\d x} {1 - a \paren {e^{i x} + e^{-i x} } + a^2}\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds -\int_0^{2 \pi} \frac {e^{i x} } {a e^{2 i x} - \paren {a^2 + 1} e^{i x} + a} \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {e^{2 i x} - \paren {a + \frac 1 a} e^{i x} + \frac a a} \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 a \int_0^{2 \pi} \frac {e^{i x} } {\paren {e^{i x} - \frac 1 a} \paren {e^{i x} - a} } \rd x\) factorising
\(\ds \) \(=\) \(\ds -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z\) Definition of Complex Contour Integral, Derivative of Exponential Function

The integrand has poles:

$z_1 = a$

and:

$z_2 = \dfrac 1 a$

We have $0 < a < 1$, so $\dfrac 1 a > 1$.

So:

$z_2$ lies outside the closed disk $\size z \le 1$

and:

$z_1$ lies in $C$

Thus the only pole of concern is $z_1$.

Therefore:

\(\ds -\frac 1 {i a} \int_{\partial C} \frac 1 {\paren {z - \frac 1 a} \paren {z - a} } \rd z\) \(=\) \(\ds -\frac {2 \pi} a \Res {\frac 1 {\paren {z - \frac 1 a} \paren {z - a} } } a\) Cauchy's Residue Theorem
\(\ds \) \(=\) \(\ds -\frac {2 \pi} a \paren {\frac 1 {2 z - \paren {a + \frac 1 a} } }_{z \mathop = a}\) Residue at Simple Pole, Derivative of Power
\(\ds \) \(=\) \(\ds -\frac {2 \pi} a \paren {\frac 1 {2 a - a - \frac 1 a} }\)
\(\ds \) \(=\) \(\ds \frac {2 \pi} {1 - a^2}\)

$\blacksquare$


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